if 2^a=3^b=6^c then show that c=ab/a+b
Answers
Note:
★ a^m × a^n = a^(m + n)
★ a^m / a^n = a^(m - n)
★ a^m × b^m = (a×b)^m
★ a^m / b^m = (a/b)^m
★ [ a^m ]^n = a^(m×n)
★ a^m = b => a = b^(1/m)
Solution:
- Given : 2^a = 3^b = 6^c
- To prove : c = ab/(a + b)
Let 2^a = 3^b = 6^c = k .
Thus,
• If 2^a = k , then
2 = k^(1/a)
• If 3^b = k , then
3 = k^(1/b)
• If 6^c = k , then
6 = k^(1/c)
Now ,
=> 6 = k^(1/c)
=> 2×3 = k^(1/c)
=> [k^(1/a)] × [k^(1/b)] = k^(1/c)
=> k^(1/a + 1/b) = k^(1/c)
=> 1/a + 1/b = 1/c
=> 1/c = 1/a + 1/b
=> 1/c = (b + a)/ab
=> c = ab/(a + b)
Hence proved .
Answer:
Given : 2^a = 3^b = 6^c
To prove : c = ab/(a + b)
Let 2^a = 3^b = 6^c = k .
Thus,
• If 2^a = k , then
2 = k^(1/a)
• If 3^b = k , then
3 = k^(1/b)
• If 6^c = k , then
6 = k^(1/c)
Now ,
=> 6 = k^(1/c)
=> 2×3 = k^(1/c)
=> [k^(1/a)] × [k^(1/b)] = k^(1/c)
=> k^(1/a + 1/b) = k^(1/c)
=> 1/a + 1/b = 1/c
=> 1/c = 1/a + 1/b
=> 1/c = (b + a)/ab
=> c = ab/(a + b)
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Step-by-step explanation: